2006-01-23T15:31:05ZFluxBBhttp://www.mathisfunforum.com/viewtopic.php?id=2576Move it, but don't delete this thread, and place a "Thread moved to here: " link for this thread.]]>http://www.mathisfunforum.com/profile.php?id=21432006-01-23T15:31:05Zhttp://www.mathisfunforum.com/viewtopic.php?pid=25429#p25429Well, I was going to move it, but then I wasn't sure whether suetonius would be able to find it again.]]>http://www.mathisfunforum.com/profile.php?id=4952006-01-23T06:48:31Zhttp://www.mathisfunforum.com/viewtopic.php?pid=25395#p25395Haha, I guess that I was thrown off by this being posted in the wrong section. It seems fitting that a wrong answer was given for a question posted in the wrong place.]]>http://www.mathisfunforum.com/profile.php?id=20412006-01-23T03:28:42Zhttp://www.mathisfunforum.com/viewtopic.php?pid=25389#p25389He was asking for the antiderivative, AKA the integral.

1/√x can be rewritten as ½x^-½. This can be integrated simply by the power rule for integration: ∫½x^-½dx = x^½ or √x.

]]>http://www.mathisfunforum.com/profile.php?id=9562006-01-23T00:58:45Zhttp://www.mathisfunforum.com/viewtopic.php?pid=25384#p25384The derivative of any fractional function:

f(x) = f(g)/f(h), f'(x) = [ f'(g)f(h) - f(g)f'(h) ] / (f(h))²

In your case;

f(g) = 1, f(h) = √x, f'(g) = 0, f'(h) = 1/(2√x)

So, using the formula for differentiating fractions above, this all becomes;

[ 0(√x) - 1(1/(2√x))] / (√x)²;

This equals;

-1/(2x√x) = (-1/2)x^(-3/2)

Note that this type of discussion belongs in the "HELP ME!" section. Please post your math questions there in the future.

]]>http://www.mathisfunforum.com/profile.php?id=20412006-01-22T21:33:16Zhttp://www.mathisfunforum.com/viewtopic.php?pid=25377#p25377Can anybody help me find the antiderivative of 1/sqrt(x)?]]>http://www.mathisfunforum.com/profile.php?id=26152006-01-22T19:30:12Zhttp://www.mathisfunforum.com/viewtopic.php?pid=25364#p25364